Here is the problem I'm trying to prove :

Let $F$ be a subset of $C([a,b])$. If each sequence in $F$ contains a uniformly convergent sequence, then $F$ is both equicontinuous and uniformly bounded.

As the assumption said, I let {$f_n$} be a sequence in $F$, so that there exists a subsequence {$f_{1,_k}$} which converges uniformly.

Since {$ {f_n} $} $\setminus$ {$f_{n_k}$} is still a sequence in $F$, so there also exists a uniformly convergent subsequence
{$f_{2,_k}$} in it.

Continue this process, I get a family of sequence {$f_{t,_k}$} with $\bigcup _t$ {$f_{t,_k}$} = {$f_n$}, and each {$f_{t,_k}$} is uniformly convergent.

Now I want to prove the result by showing that any {$f_n$} is equicontinuous and uniformly bounded, then $F$ will be. So I foucus on analyzing each {$f_{t,_k}$}. But evey method I've tried was stopped because of lacking of condition.

I don't know what I should do to keep going.

  • $\begingroup$ Have you already seen the proof of Arzela-Ascoli (which is essentially the converse of this statement)? $\endgroup$ – Ian Oct 5 '17 at 2:52
  • $\begingroup$ @Ian Yes, I have. I followed how it was proved by writhing each subsequence down, and tried to use diagonal process. But I can't find out any relation between two different {$f_{t,k}$}. $\endgroup$ – Hung Oct 5 '17 at 3:01

Hint: In fact $F \subseteq C[a,b]$ is a compact set under sup-norm so it is bounded which in language of functions means uniformly bounded.

For equicontinuouty proceed proof by contradiction and try to extract a subsequence which does not have convergent subsequence

  • $\begingroup$ Appreciate! I'll try what you said. $\endgroup$ – Hung Oct 5 '17 at 3:16

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